## anonymous one year ago check my work

1. anonymous

Q:Find $\iint_\limits S (\vec \nabla \times \vec F).\hat n \space ds$ where $\vec F=(x-z)\hat i+(x^3+yz)\hat j+3xy^2 \hat k$ and S is the surface of the cone $z=a-\sqrt{x^2+y^2}$ above the xy plane Now $\vec \nabla \times \vec F=\left|\begin{matrix}\hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\x-z & x^3+yz & 3xy^2\end{matrix}\right|$$\vec \nabla \times \vec F=\hat i(6xy-y)-\hat j(3y^2+1)+3x^2 \hat k=y(6x-1)\hat i-(3y^2+1)\hat j+3x^2 \hat k$ We can re-write our equation for cone as $\sqrt{x^2+y^2}+z=a$ Which is of the form $\phi(x,y,z)=c$ Equation of a level surface, where phi is a scalar function of x,y,z and c is a constant In this case, a vector normal to the surface is given by $\vec n=\vec \nabla \phi=(\hat i \frac{\partial \phi}{\partial x}+\hat j \frac{\partial \phi}{\partial y}+\hat k \frac{\partial \phi}{\partial z})$$\vec n=\frac{1}{2\sqrt{x^2+y^2}}.2x \hat i+\frac{1}{2\sqrt{x^2+y^2}}.2y \hat j+\hat k=\frac{x \hat i}{\sqrt{x^2+y^2}}+\frac{y \hat j}{\sqrt{x^2+y^2}}+\hat k$ $|\vec n|=\sqrt{\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}+1^2}=\sqrt{\frac{x^2+y^2}{x^2+y^2}+1}=\sqrt{1+1}=\sqrt{2}$ $\hat n=\frac{\vec n}{|\vec n|}=\frac{1}{\sqrt{2}}(\frac{x \hat i}{\sqrt{x^2+y^2}}+\frac{y \hat j}{\sqrt{x^2+y^2}}+\hat k)$ When $z=0$ We have $x^2+y^2=a^2$ This is the projection of the cone in xy-plane (circle of radius a in the xy-plane), |dw:1443798608113:dw| Thus we convert our surface integral... $\iint_\limits S (\vec \nabla \times \vec F). \hat n \space ds=\iint_\limits {R}(\vec \nabla \times \vec F). \hat n \space \frac{dxdy}{\hat n.\hat k}$ where R is the region covered by circle of radius a $\therefore \hat n=\frac{1}{\sqrt{2}}(\frac{x}{a}\hat i+\frac{y}{a}\hat j+\hat k)=\frac{1}{\sqrt{2}a}(x \hat i+y \hat j+a \hat k)$ Now $(\vec \nabla \times \vec F).\hat n=\frac{1}{\sqrt{2}a}(y(6x^2-x)-(3y^3+y)+3ax^2)$ Also $\hat n . \hat k=\frac{1}{\sqrt{2}} \implies \frac{1}{\hat n . \hat k}=\sqrt{2}$$(\vec \nabla \times \vec F).\hat n \frac{1}{\hat n . \hat k}=\frac{1}{a}(y(6x^2-x)-(3y^3+y)+3ax^2)$$\frac{(\vec \nabla \times \vec F).\hat n}{\hat n . \hat k}=\frac{1}{a}(y(6x^2-x-1)-3y^3+3ax^2)$ Thus our surface integral is $\therefore \frac{1}{a}\iint_\limits R [y(6x^2-x-1)-3y^3+3ax^3]dydx$ Using a short hand notation for the left side $\iint_\limits R =\frac{1}{a} \int\limits_{-a}^{a}[\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}((6x^2-x-1)y-3y^3+3ax^2)dy]dx$ $\iint_\limits R=\frac{1}{a}\int\limits_{-a}^{a}[(6x^2-x-1)\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}ydy-3\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}y^3dy+3ax^2\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}dy]dx$ $\iint_\limits R=\frac{1}{a}\int\limits_{-a}^{a}[0+0+6ax^2\int\limits_{0}^{\sqrt{a^2-x^2}}dy]dx=\frac{1}{a}\int\limits_{-a}^{a}6ax^2\sqrt{a^2-x^2}dx$ $\iint_{R}=6\int\limits_{-a}^{a}x^2{\sqrt{a^2-x^2}}dx=12\int\limits_{0}^{a}x^2\sqrt{a^2-x^2}dx$$x=a\sin(\theta)$$\therefore dx=a\cos(\theta)d \theta$ The limits transform as $0 \rightarrow 0 \space , \space a \rightarrow \frac{\pi}{2}$ $\iint_\limits R=12 \int\limits_{0}^{\frac{\pi}{2}}a^2\sin^2(\theta).a^2\cos^2(\theta)d \theta=12a^4\int\limits_{0}^{\frac{\pi}{2}}(\sin(\theta)\cos(\theta))^2d \theta$ $\iint_\limits R=3a^4 \int\limits_{0}^{\frac{\pi}{2}}(2\sin(\theta)\cos(\theta))^2d \theta=3a^4\int\limits_{0}^{\frac{\pi}{2}}\sin^2(2 \theta)d \theta$$\iint_\limits R=\frac{3a^4}{2}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta=\frac{3a^4}{2}[\theta-\frac{\sin(4 \theta)}{4}]_{0}^{\frac{\pi}{2}}$$\iint_\limits R=\frac{3a^4}{2}[\frac{\pi}{2}-\frac{\sin(2\pi)}{4}-(0+\frac{\sin(0)}{4})]=\frac{3a^4}{2}(\frac{\pi}{2}+0+0+0)=\frac{3 \pi a^4}{4}$

2. Jhannybean

You must've got an integration error somewhere right? :P

3. anonymous

Yikes!! Don't say such scary stuff, I miss the questions in school! Questions like these in college take a lot of time for a single question :(

4. Michele_Laino

I have computed the integral in a different way, and I got the same result

5. anonymous

sweet :D